Simple Interest for SSC MTS — Formula, Tricks, and PYQ Solutions

beginner 18 min read

Concept

Simple Interest is the most stripped-down form of interest calculation — the interest earned each year stays the same, calculated always on the original principal. Nothing compounds. Nothing snowballs. That is both its simplicity and its defining characteristic.

Think of it this way: your friend borrows ₹1000 from you at 10% simple interest per year. Every year he owes you exactly ₹100 in interest — not more, not less. After 3 years the total interest is ₹300, and the total amount he returns is ₹1300. The interest does not earn further interest. That is the key distinction from compound interest (चक्रवृद्धि ब्याज).

In contrast, if this were compound interest, the second year's interest would be calculated on ₹1100 (not ₹1000), making it slightly larger each year. Simple Interest (साधारण ब्याज) ignores that — it always looks back at the original principal.

Why does this matter for SSC MTS? Because SI questions here are nearly always straightforward substitution into the formula, or a mild twist like finding principal from interest, or dealing with changing rates across different time periods. The exam does not throw curveballs — it tests whether you remember the formula and can avoid arithmetic errors under pressure. Five questions from 2024 alone were direct applications of SI = PRT/100. Know this formula cold, know how to reverse it to find P, R, or T, and you have claimed those marks before most candidates finish reading the question.


Deep Dive

The Core Formula

SI=P×R×T100SI = \frac{P \times R \times T}{100}

Where:

The Amount (कुल राशि) returned at the end is:

A=P+SI=P+P×R×T100=P(1+RT100)A = P + SI = P + \frac{P \times R \times T}{100} = P\left(1 + \frac{RT}{100}\right)

Reversing the Formula

You will often be given SI and asked to find P, R, or T. Just cross-multiply:

P=SI×100R×TP = \frac{SI \times 100}{R \times T}

R=SI×100P×TR = \frac{SI \times 100}{P \times T}

T=SI×100P×RT = \frac{SI \times 100}{P \times R}

These three reversal forms appear constantly in SSC MTS. Do not rederive them on paper during the exam — memorise all four forms (including the original) right now.

Percentage-of-Principal Shortcut

When the question says "increase its value by 30%" and asks for time at a given rate, translate directly:

SI = 30% of P, so:

P×R×T100=30P100\frac{P \times R \times T}{100} = \frac{30P}{100}

The P cancels:

R×T=30    T=30RR \times T = 30 \implies T = \frac{30}{R}

This is a general rule: SI as a percentage of P = R × T. Memorise this. It removes the need to substitute any actual rupee value.

Multi-Rate Problems

Some questions (like the 12-year, three-rate PYQ below) give different rates for different time periods. The principle is additive — calculate SI separately for each period using the same principal, then add.

SItotal=P100(R1T1+R2T2+R3T3)SI_{total} = \frac{P}{100}\left(R_1 T_1 + R_2 T_2 + R_3 T_3\right)

Look at the structure: factor out P/100 first, then sum the R×T products. This saves you from doing three separate multiplications with the full principal.

For the 12-year question: P/100 × (8×4 + 10×3 + 15×5) = 300 × (32 + 30 + 75) = 300 × 137 = ₹41,100.

Doubling / Tripling / Becoming N Times

This is a classic SSC question type. Here's the logic:

Combining: if it doubles in t years, R = 100/t. To become N times: T = (N-1) × t.

The ratio rule: time to become N times = (N-1) × time to double. This is the fastest path through these problems.

Rate Difference Problems

"If the rate were 5% higher, interest would be ₹X more" — here you are finding principal from an extra SI caused by an extra rate. Extra SI = P × ΔR × T / 100. Solve for P directly.


Memory Tricks & Shortcuts

patternPRT Table — Never Forget the Formula

Write the formula as a triangle with SI at the top, P×R×T at the bottom (like the Distance-Speed-Time triangle). Cover whatever you want to find:

  • Cover SI → P × R × T / 100
  • Cover P → SI × 100 / (R × T)
  • Cover R → SI × 100 / (P × T)
  • Cover T → SI × 100 / (P × R)

Drawing this takes 3 seconds on your rough sheet. It eliminates wrong-formula errors entirely. Standard recall time: 8-10 seconds. With the triangle visual: 2 seconds. You gain 4-5 free seconds per question.

patternN-Times Ratio Rule

If a sum doubles in d years at SI, it becomes N times in (N-1) × d years.

Why: doubling means SI = P → rate × d = 100. For N times, SI = (N-1)P → time = (N-1) × 100/rate = (N-1) × d.

Example (the exact 2017 PYQ): doubles in 5 years → becomes 8 times in (8-1) × 5 = 35 years.

Standard method (find rate, substitute, solve): 45 seconds. This ratio rule: under 10 seconds.

eliminationCancel P When SI is a Percentage

When the question says "increases by X%" and asks for time at rate R%, set up R × T = X and solve directly. The principal is irrelevant and cancels out.

Example: "increases by 30% at 4% per annum" → T = 30/4 = 7.5 years. No rupee amount needed at all.

Standard method (assume P = ₹100, compute, verify): 40 seconds. Direct cancellation: 8 seconds.

patternFactor Out P/100 for Multi-Rate Problems

When you have 3 different rates over 3 periods, write: SI = (P/100) × (R₁T₁ + R₂T₂ + R₃T₃).

First compute the bracket: add all R×T products mentally. Then multiply by P/100 once.

Example: P = ₹30,000, periods: (8%×4) + (10%×3) + (15%×5) = 32 + 30 + 75 = 137. Then 300 × 137 = ₹41,100.

Standard method (three separate calculations then addition): 90 seconds. Factored method: 30 seconds. Saves 2 full steps.

substitutionRate Difference — Isolate P Immediately

"If rate were R% higher for T years, interest would be ₹X more" → Extra SI = P × ΔR × T / 100 = X. Isolate P in one step: P = X × 100 / (ΔR × T).

Example: ΔR = 5%, T = 4 years, extra SI = ₹5200 → P = 5200 × 100 / (5 × 4) = 5200 × 100 / 20 = ₹26,000.

This is a one-line calculation. Students who try to set up two separate interest equations and subtract take 3-4 minutes. This method: under 20 seconds.


Fast-Solving Framework

Read the question and classify it in the first 5 seconds:

Type 1 — Direct SI or Amount: Values of P, R, T all given. Plug into SI = PRT/100. Done in 15 seconds.

Type 2 — Find P, R, or T: SI given, two of the three variables given. Use the reversed formula. Done in 20 seconds.

Type 3 — Percentage increase: "Increases by X% at R%" — set R × T = X, solve for the unknown. P is irrelevant.

Type 4 — N-times problem: "Doubles in d years, becomes N times in how many?" — answer is (N-1) × d years. Do not find the rate.

Type 5 — Multi-rate: Factor out P/100, sum all R×T products in the bracket, multiply once.

Type 6 — Rate difference: Extra SI = P × ΔR × T / 100. Isolate P or whatever is unknown.

If you cannot classify within 5 seconds, write SI = PRT/100 on rough paper and identify what is known and unknown. The formula does the rest.


Solved PYQs

Why this question: The classic "becomes N times" twist that trips up candidates who try to find the rate first.

Previous Year Questionपिछले वर्ष का प्रश्न2017
A certain amount double in 5 years, when invested at simple interest. In how many years will it become 8 times?
  1. 30 years
  2. 35 years
  3. 45 years
  4. 40 years
Solutionसमाधान
Doubling in 5 years means interest = P in 5 years, so rate = 20% per annum. To become 8 times, interest needed = 7P. Time = 7P/(P×20%) = 35 years.

Solving path: Doubles in 5 years → SI = P in 5 years. Use N-times ratio rule: to become 8 times, time = (8-1) × 5 = 35 years. Answer: 35 years.


Why this question: A direct reversal — given SI, R, T, find P. Tests whether you know to reverse the formula cleanly.

Previous Year Questionपिछले वर्ष का प्रश्न2024
What sum (in ₹) will earn a simple interest of ₹240 in 4 years at 20% per year rate of interest?
  1. 150
  2. 300
  3. 600
  4. 450
Solutionसमाधान
Using SI = P×R×T/100: 240 = P×20×4/100 → 240 = 80P/100 → P = 240×100/80 = ₹300.

Solving path: P = SI × 100 / (R × T) = 240 × 100 / (20 × 4) = 24000 / 80 = ₹300. Answer: ₹300.


Why this question: Pure forward application of the formula. A gift question — do not drop it to arithmetic errors.

Previous Year Questionपिछले वर्ष का प्रश्न2024
Find the simple interest (in ₹) on ₹2500 as a sum borrowed at 4% per year rate of interest for 6 years.
  1. 640
  2. 560
  3. 540
  4. 600
Solutionसमाधान
SI = P×R×T/100 = 2500×4×6/100 = 60000/100 = ₹600.

Solving path: SI = 2500 × 4 × 6 / 100 = 60000 / 100 = ₹600. Answer: ₹600.


Why this question: Multi-rate over 12 years — the most complex SI variant in SSC MTS. Factoring P/100 is key.

Previous Year Questionपिछले वर्ष का प्रश्न2024
A person lent a sum of ₹30,000 on simple interest for 12 years, with an interest rate of 8% per annum for the first 4 years, 10% per annum for the next 3 years, and 15% per annum for last 5 years. How much interest (in ₹) will he earn at the end of 12 years?
  1. 41100
  2. 38000
  3. 40200
  4. 40800
Solutionसमाधान
Total SI = 30000×(8×4 + 10×3 + 15×5)/100 = 300×(32+30+75) = 300×137 = ₹41,100.

Solving path: Factor: SI = (30000/100) × (8×4 + 10×3 + 15×5) = 300 × (32 + 30 + 75) = 300 × 137 = ₹41,100. Answer: ₹41,100.


Why this question: "Increase by X%" question — P cancels, T = X/R is the one-step solution.

Previous Year Questionपिछले वर्ष का प्रश्न2024
How long (in years) will it take for a sum of money invested at 4% per annum on simple interest to increase its value by 30%?
  1. 5(1/2)
  2. 8(1/2)
  3. 7(1/2)
  4. 4(1/2)
Solutionसमाधान
Using SI formula, 30 = (P × 4 × T)/100 × (100/P), so T = 30/4 = 7.5 = 7(1/2) years.

Solving path: SI = 30% of P, Rate = 4%. So 4 × T = 30, giving T = 7.5 = 7½ years. Answer: 7½ years.


Why this question: Rate-difference problem — tests whether you can set up the extra-SI equation without confusing yourself with two separate interest calculations.

Previous Year Questionपिछले वर्ष का प्रश्न2024
A sum was deposited at a certain rate of simple interest for 4 years. If it had been a 5% higher rate of interest, then it would have amounted to ₹5,200 more. The sum deposited is:
  1. ₹13,000
  2. ₹10,400
  3. ₹26,000
  4. ₹25,000
Solutionसमाधान
Extra interest = P × 5% × 4 = 5200, so P × 0.20 = 5200, giving P = ₹26,000.

Solving path: Extra SI = P × 5 × 4 / 100 = 5200P × 20/100 = 5200P = 5200 × 100/20 = ₹26,000. Answer: ₹26,000.


Why this question: Another direct forward application — checks basic formula recall at a slightly different P and R combination.

Previous Year Questionपिछले वर्ष का प्रश्न2024
Find the simple interest (in ₹) on ₹2000 as a sum borrowed at 8% per year rate of interest for 5 years.
  1. 840
  2. 760
  3. 800
  4. 740
Solutionसमाधान
SI = P×R×T/100 = 2000×8×5/100 = 80000/100 = ₹800.

Solving path: SI = 2000 × 8 × 5 / 100 = 80000 / 100 = ₹800. Answer: ₹800.


Why this question: Identical structure to the ₹2000 question above, with P = ₹2500. Quick pattern recognition saves 10 seconds.

Previous Year Questionपिछले वर्ष का प्रश्न2024
Find the simple interest (in ₹) on ₹2500 as a sum borrowed at 8% per year rate of interest for 5 years.
  1. 1000
  2. 960
  3. 1040
  4. 940
Solutionसमाधान
SI = Principal × Rate × Time / 100 = 2500 × 8 × 5 / 100 = ₹1000.

Solving path: SI = 2500 × 8 × 5 / 100 = 100000 / 100 = ₹1000. Answer: ₹1000. Note: same R and T as previous question, different P — recognise the structure immediately and scale.


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